Let $f(x)=x\ln(x^2)$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{x}$ (Choice B) B $\ln(x^2)+\dfrac{1}{x}$ (Choice C) C $\ln(x^2)+2$ (Choice D) D $\ln(x^2)+1$
Solution: $f$ is a product of a function and a composite function. Let... $u(x)=x$ $v(x)=\ln(x)$ $w(x)=x^2$... then $f(x)=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $f'(x)$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=1$ $v'(x)=\dfrac{1}{x}$ $w'(x)=2x$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=1\cdot{\ln(x^2)}+x\cdot{\dfrac{1}{x^2}}\cdot2x \\\\ &=\ln(x^2)+\dfrac{2x^2}{x^2} \\\\ &=\ln(x^2)+2 \end{aligned}$ In conclusion, $f'(x)=\ln(x^2)+2$.